No. $\rm HOD$ is a model of $\sf ZFC$, therefore it has sets of reals that it thinks are not Baire measurable.
Moreover, since $M[G]$ was obtained by the Levy collapse, $\mathrm{HOD}^{M[G]}=\mathrm{HOD}^M$. So unless $M$ itself was a Solovay model, there's little to no chance that the argument will go through.
In particular, if you start with a nice ground model like $L$, then you can't get any "deeper" and $\rm HOD(HOD)$ is a moot construction.